We consider the problem of learning the optimal threshold policy for control problems. Threshold policies make control decisions by evaluating whether an element of the system state exceeds a certain threshold, whose value is determined by other elements of the system state. By leveraging the monotone property of threshold policies, we prove that their policy gradients have a surprisingly simple expression. We use this simple expression to build an off-policy actor-critic algorithm for learning the optimal threshold policy. Simulation results show that our policy significantly outperforms other reinforcement learning algorithms due to its ability to exploit the monotone property.In addition, we show that the Whittle index, a powerful tool for restless multi-armed bandit problems, is equivalent to the optimal threshold policy for an alternative problem. This observation leads to a simple algorithm that finds the Whittle index by learning the optimal threshold policy in the alternative problem. Simulation results show that our algorithm learns the Whittle index much faster than several recent studies that learn the Whittle index through indirect means.

We consider the problem of learning the optimal threshold policy for control problems. Threshold policies make control decisions by evaluating whether an element of the system state exceeds a certain threshold, whose value is determined by other elements of the system state. By leveraging the monotone property of threshold policies, we prove that their policy gradients have a surprisingly simple expression. We use this simple expression to build an off-policy actor-critic algorithm for learning the optimal threshold policy. Simulation results show that our policy significantly outperforms other reinforcement learning algorithms due to its ability to exploit the monotone property.In addition, we show that the Whittle index, a powerful tool for restless multi-armed bandit problems, is equivalent to the optimal threshold policy for an alternative problem.

Now let's do the forward pass using simple basic operations Now if we want to check our forward pass using Digraph then we can simply do it using the function that we have already created. After this code, we will get a very easy-to-understand graph, even with just the below anyone can easily understand what is happening below the table. Now, as we are done with forward pass here comes the turn of the backward pass. Before that let's add a backward function into our Digraph code and again visualize the graph. Let's visualize the graph again As soon as we hear the word backward pass then our mind already knew that we are going to deal with backpropagation.

If we do this by using gradient ascent on the log-likelihood function, each step of gradient ascent involves an expensive expectation estimate using MCMC (or some cheaper approximation). Conceptually the edge weights represent the "interaction strength" between variables, i.e. $w_{ij}$ represents how much $x_i$ and $x_j$ "want" to be equal. Just looking at the above we can see that when $w_{ij}$ is large and positive, $x_i$ and $x_j$ have a high probability of being equal and the when it's negative they have a higher probability of being opposite sign. What is the relationship between the empirical correlation between each $x_i$ and $x_j$ versus the optimal edge weight $w_{ij}$? It would make sense that variables that are highly positively correlated have large positive edge weights, and variables that are negatively correlated have negative edge weights.